answer:Since the function f(x) satisfies f(2-x) = f(x), we know that the graph of f(x) is symmetric about the line x = 1. As f(x) is an odd function over mathbb{R}, we have f(2-x) = -f(x-2) and f(x-4) = -f(4-x). Substituting x with x-2 in f(2-x) = f(x), we obtain f(2-(x-2)) = f(x-2) implies f(4-x) = f(x-2). Thus, we have f(x) = f(2-x) = -f(4-x) = f(x-4). This leads to f(x+4) = f(x), signifying that f(x) is a periodic function with a period of 4. Considering one period of f(x), for example [-1,3], and knowing that f(x) is decreasing on [0,1), it is increasing on (1,2] by the symmetry property. Similarly, f(x) is decreasing on (-1,0] and increasing on [2,3). For an odd function f(x), we have f(0) = 0 and f(2) = f(2-2) = f(0) = 0. Therefore, when x in (0,1), f(x) < f(0) = 0; and when x in (1,2), f(x) < f(2) = 0. Similarly, when x in (-1,0), f(x) > f(0) = 0; and when x in (2,3), f(x) > f(2) = 0. The equation f(x) = -1 has a real root in [0,1), which must be unique due to the monotonous nature of f(x) in (0,1). By symmetry (f(2-x) = f(x)), the equation f(x) = -1 also has a unique real root in (1,2). There are no real roots in (-1,0) and (2,3) as f(x) > 0 in those intervals. Thus, the equation f(x) = -1 has exactly two real roots within one period. When x in [-1,3], the sum of the two real roots of f(x) = -1 is x + 2 - x = 2. Extending to the interval x in [-1,7], which covers two periods, the sum of all four real roots of f(x) = -1 is x + 2 - x + 4 + x + 4 + 2 - x = 2 + 8 + 2 = 12. Therefore, the correct answer is: boxed{A}

answer:To solve for tan 105^circ, we use the angle addition formula for tangent: [ tan 105^circ = tan (60^circ + 45^circ) ] [ = frac{tan 60^circ + tan 45^circ}{1 - tan 60^circ tan 45^circ} ] [ = frac{sqrt{3} + 1}{1 - sqrt{3} cdot 1} ] [ = frac{sqrt{3} + 1}{1 - sqrt{3}} ] To simplify the expression, multiply numerator and denominator by the conjugate of the denominator: [ = frac{(sqrt{3} + 1)(1 + sqrt{3})}{(1 - sqrt{3})(1 + sqrt{3})} ] [ = frac{sqrt{3} + 1 + 3 + sqrt{3}}{1^2 - (sqrt{3})^2} ] [ = frac{2sqrt{3} + 4}{1 - 3} ] [ = frac{2sqrt{3} + 4}{-2} ] [ = -(sqrt{3} + 2) ] [ = boxed{-sqrt{3} - 2}. ]

answer:To find the intersection A cap B of the sets A={-1,0,1} and B={0,1,2}, we look for elements that are present in both sets. - The element -1 is in set A but not in set B. - The element 0 is in both set A and set B. - The element 1 is also in both set A and set B. - The element 2 is in set B but not in set A. Therefore, the intersection of sets A and B, which consists of elements common to both sets, is {0,1}. Thus, the correct answer is boxed{A}.

answer:Solution: (1) When proving by contradiction, we assume the proposition to be false, which should be a complete negation. Therefore, the false proposition for p+qleq2 should be p+q>2. Assume p+q>2, then p>2-q, p^3>(2-q)^3, p^3+q^3>8-12q+6q^2, Since p^3+q^3=2, Then 2>8-12q+6q^2, which means q^2-2q+1<0, Therefore, (q-1)^2<0, Since for any value of q, (q-1)^2 is always greater than or equal to 0, This means the assumption does not hold, Therefore, p+qleq2. Hence, the answer is p+q>2, (q-1)^2<0. This problem is solved by using the method of contradiction and analysis through definitions and scaling. This question mainly examines the definition and application of the method of contradiction and is a basic problem. Thus, the assumption for contradiction is boxed{p+q>2}, and the contradiction obtained is boxed{(q-1)^2<0}.